Question: Is ${325177}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {325177}= &&{3}\cdot100000+ \\&&{2}\cdot10000+ \\&&{5}\cdot1000+ \\&&{1}\cdot100+ \\&&{7}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {325177}= &&{3}(99999+1)+ \\&&{2}(9999+1)+ \\&&{5}(999+1)+ \\&&{1}(99+1)+ \\&&{7}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {325177}= &&\gray{3\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {3}+{2}+{5}+{1}+{7}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${325177}$ is divisible by $9$ if ${ 3}+{2}+{5}+{1}+{7}+{7}$ is divisible by $9$ Add the digits of ${325177}$ $ {3}+{2}+{5}+{1}+{7}+{7} = {25} $ If ${25}$ is divisible by $9$ , then ${325177}$ must also be divisible by $9$ ${25}$ is not divisible by $9$, therefore ${325177}$ must not be divisible by $9$.